The refcount value returned by this function is
non-obvious in certain circumstances. For example, a developer might
expect the above example to indicate a refcount of
2. The third reference is created when actually
calling debug_zval_dump.

This behavior is further compounded when a variable is not passed to
debug_zval_dump by reference. To illustrate, consider
a slightly modified version of the above example:

<?php$var1 = 'Hello World';$var2 = '';

$var2 =& $var1;

debug_zval_dump($var1); // not passed by reference, this time?>

The above example will output:

string(11) "Hello World" refcount(1)

Why refcount(1)? Because a copy of $var1 is
being made, when the function is called.

This function becomes even more confusing when a
variable with a refcount of 1 is
passed (by copy/value):

<?php$var1 = 'Hello World';

debug_zval_dump($var1);?>

The above example will output:

string(11) "Hello World" refcount(2)

A refcount of 2, here, is extremely
non-obvious. Especially considering the above examples. So what's
happening?

When a variable has a single reference (as did $var1
before it was used as an argument to debug_zval_dump),
PHP's engine optimizes the manner in which it is passed to a function.
Internally, PHP treats $var1 like a reference (in that
the refcount is increased for the scope of this
function), with the caveat that if the passed reference
happens to be written to, a copy is made, but only at the moment of
writing. This is known as "copy on write."

So, if debug_zval_dump happened to write to its sole
parameter (and it doesn't), then a copy would be made. Until then, the
parameter remains a reference, causing the refcount to
be incremented to 2 for the scope of the function call.